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Willard's General Topology says:

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if A is finite.

But, if $X_{\alpha}=\{1\}$ for each $\alpha\in A$, then $\prod_{\alpha\in A}X_{\alpha}$ is discrte space, right?

Silent
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2 Answers2

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The question has been phrased badly. It should say:

Let $A$ be a set. Then $A$ is finite if and only if for all collections $(X_\alpha)_{\alpha\in A}$ of discrete topological spaces, the product $\Pi_{\alpha \in A} X_\alpha$ is discrete.

As your example shows, moving the 'for all $(X_\alpha)$' quantifier outside gives us a sentence that is untrue.

For one direction, it should be sufficient to consider the case that $X_\alpha=\{0,1\}$ for all $\alpha$.

John Gowers
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  • This makes the statement more about the index set than about products of discrete spaces. – Henno Brandsma Apr 17 '16 at 18:26
  • Why $X_{\alpha}={1}$ does not create problem in your statement? Isn't $\Pi_{\alpha \in A} X_\alpha$ singleton again? – Silent Apr 17 '16 at 18:34
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    @Silent The reverse implication in my statement is that if $\Pi X_\alpha$ is discrete for all collections $X_\alpha$ of discrete spaces, then $A$ is finite. If $A$ is infinite, there may be particular cases in which $\Pi X_\alpha$ is discrete - such as when $X_\alpha={1}$ for all $\alpha$, but there must be at least one collection of discrete $X_\alpha$ such that $\Pi X_\alpha$ is not discrete. Henno's answer gives a stronger statement - the case when $\Pi X_\alpha$ is discrete is precisely the case when all but finitely many $X_\alpha$ have one point. – John Gowers Apr 17 '16 at 18:53
  • I would put the products of discrete spaces first, but it's a matter of taste, I suppose. Your statement is perfectly valid, I concur. – Henno Brandsma Apr 17 '16 at 18:55
  • @HennoBrandsma I would too, but it turned out to be remarkably difficult to write the sentence that way without the order of quantifiers being ambiguous. I completely agree that it's not ideal. – John Gowers Apr 17 '16 at 19:09
  • @Donkey_2009, thank you so much for comments. I need just one clarification : you are not fixing $(X_\alpha){\alpha\in A}$, that is $\Pi{\alpha \in A} X_\alpha$ is arbitrary, right? – Silent Apr 18 '16 at 07:28
  • That is right. $\quad$ – John Gowers Apr 18 '16 at 11:32
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You're right that the formulation is not optimal.

I think he meant to say:

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space with more than one point. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if A is finite.

Or

For each $\alpha\in A$, let $X_{\alpha}$ be discrete topological space. Then $\prod_{\alpha\in A}X_{\alpha}$(under product topology) will be a discrete space if and only if $\{\alpha \in A: |X_\alpha| > 1\}$ is finite.

Or as an escape clause: maybe Willard defines somewhere that a discrete space is by definition one that has at least 2 points, or some such trick. I don't have it at hand now.

In simple terms: he wants to state that "all" infinite products of discrete spaces are no longer discrete any more. But he does need to add the clause that the spaces are not singletons to avoid trivialities.

Henno Brandsma
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  • The second formulation is still not correct, because the product will be discrete if even a single one of the factors is empty. – Eric Wofsey Jul 17 '16 at 03:47
  • @EricWofsey but Willard assumes spaces are non-empty by definition. I think I even found somewhere in the book that a discrete space has to have at least 2 points (which is a bit strange, admittedly). – Henno Brandsma Jul 17 '16 at 04:50